# Find area of triangle given one angle and the lengths its altitude divides the opposite side into

In triangle $$\triangle ADC$$, $$DB$$ is perpendicular to $$AC$$ at $$B$$ so that $$AB=2$$ and $$BC=3$$ as shown in the figure. Furthermore, $$\angle ADC=45^\circ$$ . Find the area of $$\Delta ADC$$.

Tried the following things:

1. Drawing medians and altitudes of the right triangles formed.
2. Apollonius's theorem.
3. Stewart's theorem.
4. Sine rule.
5. Cosine rule.
6. Noticing that $$3+2$$ equals to $$5$$, I extended segment $$BD$$ to form a 3-4-5 triangle and try to proceed.

Spent around 1.5 hours on it. In each of the cases, I wasn't able to proceed after a certain point.

• Commented Aug 20, 2023 at 15:49

$$1=\tan(\alpha + \beta) = {\tan(\alpha)+\tan(\beta)\over 1-\tan(\alpha)\tan(\beta)}$$

$${\tan(\alpha) \over \tan(\beta)} = {2\over 3}$$

Solve them you get $$\tan(\beta)={1\over 2}$$ so the height is $$6$$

• I was shocked by seeing such a simple solution to the problem i spent 1.5 hr on. Thanks A Lot Commented Aug 20, 2023 at 16:13

In a triangle $$\,\triangle ADC\,,\,$$ $$DB\,$$ is perpendicular to $$\,AC\,$$ at $$B\,$$ so that $$\,AB=2\,$$ and $$\,BC=3\,$$ as shown in the figure. Furthermore, $$\,\angle ADC=45\unicode{176}\,.$$
Find the area of $$\,\triangle ADC\,.$$

Here is a solution which does not use trigonometric functions.

First of all we consider the circumscribed circle of $$\,\triangle ADC\,.$$

Since a central angle of a circle is twice any inscribed angle subtended by the same arc, it follows that

$$\angle AOC=2\angle ADC=2\!\cdot45\unicode{176}=90\unicode{176}.$$

So $$\,\triangle AOC\,$$ is a rectangle and isosceles triangle, hence it is a half-square triangle and consequently

$$OA=OC=\dfrac{AC}{\sqrt2}=\dfrac5{\sqrt2}\,.\quad\left(\implies OD=\dfrac5{\sqrt2}\right)$$

Moreover, $$\,\triangle OCH\cong\triangle AOH\,$$ are two congruent half-square triangles, therefore it results that

$$CH=OH=HA=\dfrac{AC}2=\dfrac52\,.$$

On the other hand, since $$\,OHBK\,$$ is a rectangle, we get that

$$KB=OH=\dfrac52\;\;,$$

$$OK=HB=BC-CH=3-\dfrac52=\dfrac12\;\;,$$

and, by applying Pythagoras’ theorem, it results that

$$DK=\sqrt{OD^2-OK^2}=\sqrt{\left(\dfrac5{\sqrt2}\right)^2-\left(\dfrac12\right)^2}=\dfrac72\;.$$

Therefore ,

$$DB=DK+KB=\dfrac72+\dfrac52=6\;\;,$$

and the area of $$\,\triangle ADB\,$$ is equal to

$$\mathrm{Area}=\dfrac{AC\cdot DB}2=\dfrac{5\cdot6}2=15\;.$$

• Beautiful! Very clever to use the double angle at the center of the circumcircle! Commented Aug 21, 2023 at 13:01

With the cosine law:

Set $$h = BD$$. We have

$$\frac{\sqrt{2}}{2}=\cos(45^o) = \frac{DC^2+AD^2-CA^2}{2\ DC\ AD} = \frac{3^2+h^2+2^2+h^2-5^2}{2\sqrt{3^2+h^2}\sqrt{2^2+h^2}}$$

and solve for $$h$$ to get $$h=6$$. (One has to be careful: squaring the equations introduces an extraneous solution, which one needs to discard).

Rotate the triangle and draw it on the complex plane, so $$z_A=h+2i$$, $$z_B=h$$, $$z_C=h-3i$$, and $$z_D=0$$, where $$h > 0$$ is a real number representing the length of $$BD$$.

I will work in radian measure so $$45°$$ is $$\pi/4$$ radians. This is the difference in argument between $$z_A$$ and $$z_D$$, so we have

\begin{align} \arg(h+2i) - \arg(h-3i) &= \frac{\pi}{4} \\ \arg \left( \frac{h+2i}{h-3i} \right) &= \frac{\pi}{4} \tag{1} \\ \arg \left( \frac{(h+2i)(h+3i)}{(h-3i)(h+3i)} \right) &= \frac{\pi}{4} \tag{2} \\ \arg \left( \frac{(h^2-6) + (5h)i}{h^2 + 9} \right) &= \frac{\pi}{4} \tag{3} \\ \arg ((h^2-6) + (5h)i) &= \frac{\pi}{4} \tag{4} \\ \frac{5h}{h^2-6} &= \tan \frac{\pi}{4} = 1 \tag{5} \\ h^2 - 6 &= 5h \\ h^2 - 5h - 6 &= 0 \\ (h-6)(h+1) &= 0 \\ \end{align}

Since we want $$h>0$$, we must take $$h=6$$, so the area of $$\triangle ADC$$ is $$\frac{1}{2}\cdot (2+3) \cdot 6=15$$.

Steps: $$(1)$$ argument of quotient of complex numbers is the difference in their arguments (at least modulo $$2 \pi$$), $$(2)$$-$$(3)$$ multiply numerator and denominator by complex conjugate of denominator to make the denominator real and positive, $$(4)$$ argument of a complex number is unchanged by multiplying or dividing it by a real, positive number since you stay on the same ray from the origin, $$(5)$$ note that if $$x+yi$$ has argument between $$-\pi/2$$ and $$\pi/2$$ then $$x>0$$ and the formula for the argument is simply $$\arctan(y/x)$$. Life is more complicated elsewhere. Without using any trigonometric functions, we could have noted that $$\arg(x+yi)=\pi/4$$ is only true for the half-line $$y=x$$, $$x>0$$, and deduced immediately that $$h^2-6=5h$$.

The above is basically equivalent to the following trigonometric argument using the arctangent addition formula:

$$\arctan{u}+\arctan{v}=\arctan\left(\frac{u+v}{1-uv}\right)$$

Note that this formula holds when $$uv < 1$$. Here if $$u = \tan \angle ADB = 2/h$$ and $$v = \tan \angle CDB = 3/h$$ then since both angles are less than $$45°$$, their tangents $$u$$ and $$v$$ will be less than one, and so will the product $$uv$$.

We have \begin{align} \arctan(2/h) + \arctan(3/h) &= 45° \\ \arctan\left( \frac{2/h + 3/h}{1 - (2/h)(3/h)} \right) &= 45° \\ \arctan\left( \frac{5/h}{1 - 6/h^2} \right) &= 45° \\ \arctan\left( \frac{5h}{h^2 - 6} \right) &= 45° \\ \end{align}

as before.